Why is population variance estimated to be $\frac{1}{N-1}\Sigma_{1 \leq i\leq N}(x_i-m)^2$ as opposed to sample variance which is $\frac{1}{N}\Sigma_{1 \leq i\leq N}(x_i-m)^2$, where m is the mean? I know that this is the unbiased estimate of population mean. But i am not able to grasp the intuition behind it. What does the term degree of freedom mean in this context?
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2This has been asked and answered more than once on MSE. Here is one of them. There have been others. I picked this one up from the right side of this page, under Related. – André Nicolas Jan 16 '15 at 00:57
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Thanks for pointing out. I will go through these answers. – Curious Jan 16 '15 at 01:08
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1Another related question: Intuitive Explanation of Bessel's Correction. I knew about this one because I had written an answer for it :-) – robjohn Jan 16 '15 at 01:27
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Thanks I was looking for an intuitive explanation of degree of freedom. i will go through the mentioned articles. – Curious Jan 16 '15 at 01:34
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The comment made by Andre addresses the majority of your question. As to the rest, what does "degrees of freedom" mean, it is literally the number of free variables when you are computing whatever you are computing. if you calculate the sample variance without subtracting 1 from $N$ then you are assuming that you have one more degree of freedom than you actually have, because you are computing a number using the entire sample, which introduces an equation which reduces the number of degrees of freedom by 1.
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If we consider the mean to be known the degree of freedom is N-1. Does the degree of freedom for calculating higher moments keeps on decreasing. – Curious Jan 16 '15 at 01:52